hdu 1250 Hat's Fibonacci

http://acm.hdu.edu.cn/showproblem.php?pid=1250

 

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3943    Accepted Submission(s): 1346

Problem Description

A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1. F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4) Your task is to take a number as input, and print that Fibonacci number.

 

Input

Each line will contain an integers. Process to end of file.

 

Output

For each case, output the result in a line.

 

Sample Input

100

 

Sample Output

4203968145672990846840663646 Note: No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

 

Author

戴帽子的

 

Recommend

Ignatius.L

 

自己写的不怎么样，用了500+MS  不过看到了一个牛B的代码，还是保存下来，有时间慢慢研究。

我的代码：

#include <stdio.h>
#include <string.h>
#define MAX 3000+3

char str1[MAX],str2[MAX],c[MAX];

void jia(char str1[],char str2[],char sum[])
{
    int i,j,k,z;
    z=0;
    k=0;
    for(i=strlen(str1)-1,j=strlen(str2)-1;i>=0||j>=0;i--,j--)
    {
        if(i>=0)
            z+=str1[i]-'0';
        if(j>=0)
            z+=str2[j]-'0';
        c[k++]=z%10+'0';
        z/=10;
    }
    if(z) c[k++]='1';
    i=0;
    for(--k;k>=0;k--)
        sum[i++]=c[k];
    sum[i]='\0';
}

int main()
{

    int i,n;
    //char str1[MAX]="0",str2[MAX]="1",f[MAX];
    while(scanf("%d",&n)==1)
    {
        char str1[MAX]="1",str2[MAX]="1",str3[MAX]="1",str4[MAX]="1",f[MAX],f1[MAX],sum[MAX];
        if(n==0)
        {
            printf("0\n");
            continue;
        }
        for(i=4;i<n;i++)
        {
            jia(str1,str2,f);
            jia(str3,str4,f1);
            jia(f,f1,sum);
            strcpy(str1,str2);
            strcpy(str2,str3);
            strcpy(str3,str4);
            strcpy(str4,sum);
        }
        printf("%s\n",str4);
    }
    return 0;
}

 

http://www.cnblogs.com/newpanderking/archive/2011/07/31/2122528.html

用二维数组模拟大数加法，每一行表述一个数，每一行的一个元素可以代表n位数，这个可以根据自己的需要自己定义。

其他的就和正常的加法一样了，注意进位处理。

代码：

#include <iostream>
#include <stdio.h>
using namespace std;
int s[7500][670];
void solve(){
    s[1][1] = 1;s[2][1] = 1;
    s[3][1] = 1;s[4][1] = 1;
    int i,j,k=0;
    for(i = 5;i<7500;i++)
    for( j = 1;j<=670 ;j++)
    {
        k += s[i-1][j]+s[i-2][j]+s[i-3][j]+s[i-4][j];
        s[i][j] = k%10000;
        k = k/10000;
    }
    while(k)
    {
        s[i][j++] = k%10000;
        k = k/10000;
    }
}
int main()
{
    int n,i,j;
    solve();
    while(cin>>n)
    {
        for(i = 670; i>=1;i--)
        if(s[n][i]!=0)break;
        printf("%d",s[n][i]);
        for(j = i-1;j>=1;j--)
        printf("%04d",s[n][j]);
        printf("\n");
    }
}